Theoretical Models of Chemical Processes

7/31/2019 Theoretical Models of Chemical Processes

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Chapter 2: Theoretical Models of Chemical Processes 1

Theoretical Models of Chemical Processes

1.Goals

By the end of this chapter, the students should be able to do the following:a. Formulate dynamic models based on fundamental balances.b. Linearize nonlinear model equations.

2. FundamentalsDefinition. A mathematical model of a process is a system of equations whose solution,

given specific input data, is representative of the response of the process to a

corresponding set of inputs.

Mathematical models can be classified based on different criteria:

1. Theoretical and empirical models.2. Steady state and dynamic models.3. Lumped and distributed models.

Mathematical models can be useful in different chemical engineering phases:

1. Research and development: determining chemical kinetic mechanisms andparameters from laboratory or pilot-plant reaction data; exploring the effects ofdifferent operating conditions for optimization and control studies; aiding in

scale-up calculations.

2. Design: evaluating alternative process and control structures and strategies;simulating start-up, shutdown and emergency situations and procedures.

3. Plant operation: aiding in operators training, optimizing plant operation.Notice that modeling is performed to answer specific questions; thus, no one model is

appropriate for all situations.

3. Modeling procedure

1. Define goals.

Goal statement is a critical element of the modeling task since it determines the type of

the model needed. The goal should be specific concerning the type of information


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needed; for example numerical values, semi quantitative information about system

characteristics.

2. Prepare information.

a. Sketch the process.

b. Collect data.

c. State assumptions. The model should be no more complicated than necessary to

meet the modeling objectives.

d. Define the system

3. Formulate the model.

1. Overall mass balance.2. Component mass balance.3. Energy balance

4. Determine the Solution.

The resulted dynamic model will involve differential (and algebraic) equations,

because of the accumulation terms, with initial conditions and some change to an input

variable, i.e. forcing function. Depending on model complexity, model solution isobtained either analytically or numerically.

5. Results analysis

The first phase is to evaluate whether the solution is correct. This can be partially

verified by considering the following:

Goal

How detailed

the model is?

What is the required

decision?

What is the required

variable?


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a. The result satisfies initial and final conditions.b. Obeys assumptions.c. Sign and shape as expected.

The second phase is the extensive study and interpretations of the obtained solution.

Finally, the sensitivity of the results to changes in assumptions or data should beevaluated (what ifanalysis).

6. Validation

Compare with empirical data.

4. Fundamental laws

Consider the system shown in Figure 1.

Figure 1. A general system and its interaction with outside world

1. Total mass balance

{Accumulation of mass} = {mass in} {mass out}

=outletj

jj

inleti

ii FFdt

Vd

::

)(

2. Mass balance on component A

{Accumulation of component mass} = {component mass in}{component mass out}

{generation/consumption of component mass}

==outletj

AjAj

inleti

iAiAA VrFcFc

dt

Vcd

dt

dn

::

)(

System

Work

Heat

Inlets

Fi,i, Ti,i

Ci,i, i,i

Oulets

Fi,o, Ti,o

Ci,o, i,o


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3. Energy balance

{Accumulation of internal, kinetic and potential energy}={flow of internal, kinetic and

potential energy into system}-{flow of internal, kinetic and potential energy out of

system}+{Heat added to the system}-{work done by the system on surrounding}

WQHFHFdt

PKUdoutletj

jjj

inleti

iii =++ ::

)(

5. Degrees of freedom in modeling

Question

Consider the following system

2515105

134

532

=++

=++

=++

yxz

yxz

yxz

Does this system have a unique solution?

Answer

No. Because, the number of variables is larger than the number of equations.

To use a mathematical model for process simulation we must insure that the model

equations, differential and algebraic, provide a unique relation among all inputs and

outputs. For a system of large, complicated steady or dynamic model, there is a unique

solution if the number of unknown variables equals the number of independent model

equations. An equivalent way of stating this condition is to require that the degrees of

freedom (DOF) be zero, that is:

DOF = NV-NE

NVrepresents the number of variables (unspecified inputs and outputs) that depend on

the behavior of the system and are to be evaluated through the model equations.

NEis the number of independent equations (both differential and algebraic). The degrees

of freedom analysis separates modeling problems into three categories:

1. DOF = 0; the system is exactly specifiedand the set of equations has a solution. Notethat this solution may not be unique for a set of nonlinear equation.

2. DOF < 0; the system is over-specified. In general no solution to the model exists. Thisis a symptom of an error in the formulation. The likely cause is either (1) considering

a variable as a parameter or (2) including an extra dependent equation(s) in the

model.


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3. DOF > 0; the system is underspecified. An infinite number of solutions to the modelexist. The likely causes are (1) considering a parameter as a variable or (2) not

including in the model all equations that determine the systems behavior.

The steps in the degrees of freedom analysis are summarized in Table 1.

Table 1. Degrees of freedom analysis

6. Examples

Example 1

The dynamic behavior of a (constant volume) mixing (blending) tank is to be modeled.

The tank has been designed well with buffering and impeller size, shape and speed such

that the concentration should be uniform in the liquid. Due to the low concentration of

component A in the solvent, it can be assumed that the density of reactor contents is

constant.

As a second task, derive model equations for a variable volume system.

Data: Fin = 0.085 m3/min; V = 2.1 m

3; CA,in = 0.925 mol/m

3.

Solution:

The goal is to study the dynamic behavior of the system. Thus, it is vital to write down

all dynamic equations that describe the system.

Assumptions:

a. Constant volume mixer.b. Mixer contents are perfectly mixed.c. Heat of mixing is negligible, i.e. mixer temperature is constant.d. Density of mixer contents is constant.


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Sketch:

Inputs: Fin and CA,in.

Parameters: V,

Outputs: F, CA

Model Equations:

Total mass balance:

FFdt

dV

dt

dV

dt

Vdin

=+=

)(

Because of assumptions (a) and (d)

FF

FFdtdV

in

in

=

== 0)(

Component A mass balance:

)()( , AinAA CCFVCdt

d=

NV = CA and F

NE = 2

DOF = NV NE = 2 2 = 0.Note, the liquid volume is constant when:

1. An overflow line is used in the tank as shown in the sketch for this example.2. The tank is closed and filled to capacity.3. A liquid-level controller keeps V constant by adjusting a flow rate.

F

V CA

Fin

CA,in


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Example 2

A typical liquid storage process is shown in the figure below. q i and q are volumetric

flow rate. There are three important variations of the liquid storage process. Discuss these

variations.

Solution:

1. The inlet or outlet might be constant. The outlet might be maintained by a constant

speed, fixed volume pump. An important consequence of this configuration is that the

outlet flow rate is completely independent of the head in the tank. The tank operates as

flow integrator.

,qqdt

dhA i = Note that right-hand side is constant

It can be proved that the solution for this equation is

tA

)qq(hh io

+=

Interestingly, liquid height in the tank might linearly decrease or increase. This will make

the tank run empty or flood respectively.

Notice that the output grows (decreases) linearly with time in unbounded fashion (the

figure below). Thus

tasy /

2. The tank exit line may function simply as a resistance to flow from the tank or it may

contain a valve that provides significant resistance to flow at a single point. The flow may

be assumed to be linearly related to the deriving force, liquid head, in analog to Ohms

low:

q = driving force for flow/resistance to flow

qi

h

q

V

A


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hR

1q

v

=

Rv is the resistance of the line. The use of this equation yields a first-order linear

differential equation:

hR

1q

dt

dhA

v

i =

The solution h(t) can be found analytically in cases where q i(t) can be specified as simple

functions of time.

3. A more realistic expression for flow can be obtained when a valve has been placed in

the exit line. The pressure difference deriving flow through the valve is

aPPP =

P is related to q,

av PPCq =

Where Cv, called valve coefficient, depends on the particular valve used and its flow

rating. Here we assume that the flow discharges at ambient pressure Pa and that the

upstream pressure P is the pressure at the bottom tank

ghPP a +=

The combination of these equations results in the following differential equation

ghCqdt

dhA vi =

The liquid storage processes discussed above could be operated by controlling the level

in the tank or by allowing the level to fluctuate without attempting to control it. In the

later case, it may be of interest to predict weather the tank would overflow or run dry forparticular variations in the inlet and outlet flow rates. Hence, the dynamics of the process

may be important even when automatic control is not utilized.

y(t)

tA

qqi


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Example 3

Consider the stirred-tank heating system shown in the Figure below. The liquid inlet

stream consists of a single component with a mass flow rate w and an inlet temperature

Ti. the tank contents are agitated and heated using an electrical heater that provides a

heating rate, Q. Discuss the development of a dynamic model for this system based onconstant liquid hold up assumption first; then discuss the effect of relaxing this

assumption on the dynamic model.

Solution.

Assumptions:

1. Perfectly mixed system.2. Liquid hold-up is constant.3. The density and heat capacity of the liquid are assumed to be constant. Thus, their

temperature dependence is negligible.

4. Heat losses are negligible.5. The net rate of work is negligible compared to the rates of heat transfer and

convection.

Case 1: constant hold up

Inputs: wi, Ti, Q

Parameters: V, C,

Outputs: w, T

Total mass balance


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wwdt

Vdi == 0

)(

Energy balance

For a pure liquid at low or moderate pressure, the internal energy is approximately equal

to the enthalpy, U H, and depends only on temperature.

[ ]QTTwC

QTTCTTCw

QHHwdt

dTVC

dt

dHm

dt

mHd

i

refrefi

outi

+=

+=

+===

)(

)()(

)()(

NV = 2, NE = 2.

DOF = NV-NE = 2-2 = 0

Case 2: variable hold-up

Total mass balance

wwdt

dV

wwdt

Vd

i

i

=

=

)(

Energy balance

QHwHw

dt

dVH

dt

dHV

dt

VHd

dt

mHd

ii +=

+== ][)()(

Note that dH/dt = CdT/dt. Substituting this equation into energy balance equation and

using mass balance give:

QTTwCTTCw

QwHHw

dt

dTCVwwTTC

dt

VHd

refrefii

ii

iref

+=

+=

+=

)()(

))(()(

Rearranging this equation gives

QTTCV

w

dt

dTi

i +

= )(


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This example has demonstrated that process models with variable hold-up can be

simplified by substituting the overall mass balance into the other conservation equations.

Inputs: wi, Ti, Q

Parameters: C, Outputs: V, T, w

DOF = 3-2 = 1 ???...

Total mass and energy balance equations provide a model that can be solved for the two

outputs (V and T) if the two parameters (C, ) are know and the four inputs (w, wi, Ti and

Q) are known functions of time.

Example 4

The system to be considered in this example is a CSTR. The reactor is fed with

components A and B with inlet concentrations CAo and CBo , in mol/m3, respectively.

Consider a simple first-order, irreversible exothermic chemical reaction where chemical

A species reacts to form species B:

BA k

Assuming constant inlet flow rate, qi, and the reactor operates with constant mass hold-

up, write the equations describing the system.

Solution:Assumptions:

Reactor is well-mixed. Constant volume reactor. Shaft work is negligible. Heat lose to the atmosphere is negligible. Enthalpy is only a function of temperature. Jacket volume is constant Heat capacity is constant. Thermal capacitance of the coil (reactor jacket) wall is negligible.

Sketch:


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Variables:

Inputs: Fin, CA,in, CB,in, in, Tin

Outputs: F, CA,, CB,, T

Parameters: k, H, U, Cp,A, Cp,B (or hA, hB),

Overall mass balance

FF

FFdt

dV

in

in

=

== 0)(

Mass balance on component A

=

=

=

RT

Ekk

kCr

VrCCFCdt

dV

o

AA

AAinAA

exp

)()( ,

We can write another balance equation for B

VrCCFCdt

dV ABinBB += )()( ,

Or it can be calculated using the following equation

=+ BwBAwA CMCM

Since the system is binary the overall mass balance and only one component continuity

equation is required.


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Energy balance

JArefpinrefp

refp

refp

in

QVrHTTCFTTCFdt

TTVCd

TTCH

Vm

WQhFhFdt

mHd

++=

=

=

++=

)())(())(())(

(

)(

)()()(

Taking the overall mass balance and the assumptions into account, the above equation

can be simplified as follows

JAinpp QVrHTTCFdt

dTVC ++= )()(

Reactor jacket:

1. In the first case coolant flow is high. Thus, its temperature will be assumed to beconstant (TJ).

)()()( TTUAVrHTTCFdt

dTVC JAinpp ++=

U is overall heat transfer coefficient

A is heat transfer area.

2. Perfectly mixed cooling jacket. We assume that the temperature everywhere in the

jacket (TJ).

Jacket energy equation

)()(

)()(

JJopJJJJ

pJJJ

JJoJJJ

JJ

TTUATTCFdt

dTCV

OR

TTUAhhFdt

dhV

+=

+=

The relationship between area and holdup (V) needs to be estimated. If the reactor is a

flat-bottomed vertical cylinder with diameter D and if the jacket is only around outside,

not around the bottom:

D

VA 4=

3. Plug flow cooling jacket. In many jacketed vessels the rise in water temperature is

significant and the coolant flow is more like plug flow than perfect mix. In this case, an

average jacket temperature (TJA) is used:


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2

JexitJOJA

TTT

+= (*)

The average temperature is used in the heat transfer equation and to represent the

enthalpy of jacket material. Jacket equation becomes:

)()( JAJexitJopJJJJApJJJ TTUATTCFdt

dTCV +=

This equation is integrated to obtain TJA at each time instant in time. Equation (*) is used

to calculate TJexit also as a function of time.

DOF

NV = 4; NE = 3

Another case

Note that in the above examples, wall temperature is assumed equal to the temperature

of the system which is in touch with. This simplifies the modeling. Consider a system

with wall that has large thermal capacitance. Furthermore, assume that the wall has a

uniform temperature, Tw.

Consider the stirred-tank system with steam heating coil. We assume that the thermal

capacitance of the liquid condensate is negligible compare to the thermal capacitances of

the tank liquid and wall of the heating coil. This is a realistic assumption as long as a

stream trap is used to remove condensate from the coil as it is produced. As a result, the

dynamic model consists of energy balances on the liquid and heating coil wall:

)()(

)()()(

TTAhTTAhdt

dTCm

TTAhVrHTTCFdt

dTVC

wppwsssw

pww

wppAinpp

=

++=

Note that it is not possible to consider the overall heat transfer coefficient; instead, heat

transfer coefficient, hp, is used. Subscripts p and w refer to the reactor side and wall

respectively. However, the system is not well-specified, what about Ts, which is steam

temperature. This temperature can be fixed when condensate pressure (Ps) is known,

because both variables are related by thermodynamic relation, i.e. Ts = f(Ps). Thus, the

dynamic model contains three temperature outputs (Ts, Tw and T). Two of them are

determined by dynamic equations and the third using an algebraic equation.


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Example 5

Suppose a mixture of gases is fed into the reactor sketched in the figure below. The

reactor is filled with reacting gases which are perfectly mixed. A reversible reaction

occurs:

AB

BAk

k

2

1

The forward reaction is 1.5th

-order in A; the reverse reaction is first order in B. The mole

fraction of reactant A in the reactor is y. the pressure inside the vessel is P. Both P and y

can vary with time. The system is assumed to be isothermal. Prefect gases are also

assumed. The feed stream has a density o and a mole fraction yo of reactant A. it is

volumetric flow rate is Fo.

The flow out of the reactor passes through a control valve into another vessel which is

held at a constant pressure PD. Flow through the control valve is calculated by

Dv

PPCF

=

With Cv is the valve sizing coefficient.

Solution:

Sketch

Assumptions:

Well-mixed system. Gas fills the whole volume of the reactor; thus, system volume can be assumed to be

constant.

Isothermal constant. Ideal gas low is applicableVariables:

Inputs: Fo, yo, T, PD, CAo, o

Outputs: F, y, , CA, P

Parameters: V, MA, MB, Cv, k1, k2, R

Fo, yo

o

V P

T y

F, y

PD


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Overall mass balance:

FFdt

dV oo

=

Component A continuity

BAAAooA CVkCVkFCCF

dtdCV 25.11 +=

Density varies with pressure and composition. It can be calculated using the ideal gas

low.

RT

P

V

n

nRTPV

=

=

Multiplying both terms by the molecular weight of the average one in case of mixtures

=

MV

n

M

RT

PM

V

n

[ ]RT

PMyyM

RT

PMBA )1( +==

M is the average molecular weight. The concentration of reactant in the reactor is:

RT

PyCA =

Example 6

Derive a mathematical model that describes the removal of sulfur dioxide (SO2) from

combustion gas by use of a three-stage absorption unit.

Solution

Assumptions:

1. Due to the extensive mixing we can assume that the component to be absorbed is inequilibrium between the gas and liquid streams leaving each stage i. a simplerelationship is usually assumed. For stage i:

baxy ii +=

2. Constant liquid hold up.3. Perfect mixing on each stage.4. Neglect gas hold up.


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5. Molar liquid and gas flow rates are unaffected by the absorption because thechanges in concentration of the absorbed component are small. Thus, both flow

rates are approximately constant.

Sketch:

Stage i

G, yi-1

G, yi

L, xi+1

L, xi

Variables:

Inputs: xf, yf, G, L

Parameters: a, b, H

Outputs: x1, x2, x3, y1, y2, y3

Balances:

Overall mass balance is useless because of assumption 5.

Component material balance in stage i, gives

3,2,1),()( 11

11

=+=

+=

+

+

ixxLyyG

LxGyLxGydt

dxH

iiii

iiiii

Substituting the equilibrium equation into this equation gives

11 )( + ++= iiii LxxaGLaGx

dt

dxH

Dividing by L and substituting = H/L (the stage liquid residence time), = aG/L (the

stripping factor), and K = G/L (the gas-to-liquid ratio), the following model is obtained

for the three-stage absorber:

f

f

xxxdt

dx

xxxdt

dx

xxbyKdt

dx

++=

++=

++=

323

3212

211

)1(

)1(

)1()(


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7. Linearization

7.1 Introduction

Linearization is a process by which we approximate nonlinear systems with linear

ones. It is widely used in the study of process dynamics and design of control systems for

the following reasons:

1. We can have analytical solutions for linear systems. Thus, we can have a completepicture and general picture of a processes behavior independently of the particular

values of the parameters and input variables. This is not possible for nonlinear

systems, and computer simulation provides us only with the behavior of the system at

specified values of inputs and parameters.

2. All the significant developments toward the design of effective control systems havebeen limited to linear processes.

The first question to be answered is just what a linear differential equation is. Basically, it

is one that contains variables only to the first power in any one term of the equation.

Linear example

)(1 tfxadt

dxa o =+

Where ao and a1 are constants or functions of time only, not of dependent variables or

their derivatives.

Nonlinear examples

)()()(

)(

)(

211

1

1

5.0

1

tftxtxadt

dxa

tfeadt

dxa

tfxadt

dxa

o

x

o

o

=+

=+

=+

Where x1 and x2 are both dependent variables.

7.2 Linearization of systems with one variable.

Consider the following nonlinear differential equation, modeling a given process:

)(xfdt

dx= (**)

Expand the nonlinear functionf(x) into a Taylor series around the point xo and take


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LL +

++

+

+=

!

)(

!2

)(

!1)()(

2

2

2

n

xx

dx

fd

xx

dx

fdxx

dx

dfxfxf

n

o

xo

n

n

o

xo

o

xo

o

If we neglect all terms of order two and higher, we take the following approximation for

the value off(x):

)()()( oxo

o xxdx

dfxfxf

+

In Figure 1 we can see the nonlinear function f(x) and its linear approximation around

xo. From the same picture it is also clear that the linear approximation depends on the

location of the point xo around which we make the expansion into a Taylor series.Compare the linear approximation off(x) at the points xo and x1. The approximation is

exact only at the point of linearization.

)()()( oxo

o xxdx

dfxfxf

+

In Equation (**), replace f(x) by its linear approximation given by the last equation

above and take

)()( oxo

o xxdx

dfxf

dt

dx

+=

Example

Consider the tank system shown in the figure below.

f(x)

f(xo)

xo x1 x

f(x)


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The total mass balance yields

FFdt

dhA i =

Consider

,hF = = constant

The resulting total balance yields a nonlinear dynamic model,

iFhdt

dhA =+

Let us develop the linearized approximation for this nonlinear model. The only nonlinear

tem is ,h taking the Taylor series expansion around a point ho:

L+

+

+=

==!2

)()(

2

2

2

o

hh

o

hh

o

hhh

dh

dhhh

dh

dhh

oo

Neglecting the terms of order two and higher, we take

)(2

1o

o

o hhh

hh +

By introducing this equation in the nonlinear dynamic system

oi

o

hFhhdt

dhA

22

=+

Let us compare the linearized approximate model to the nonlinear one. Assume that the

tank is at steady state with a liquid level ho. Then, at t = 0, the supply of liquid to the tank

is stopped, while we allow the liquid to flow out. Curve A is the solution of the nonlinear

Fi

F

h


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dynamic equation and curve B is the solution of the linearized dynamic model. Notice

that the two curves are very close to each other for a certain period of time. This indicates

that the linearized model approximates well the nonlinear model at the beginning. As the

time increases and the liquid level continues to fall, its value h diverts more and more

from the initial value around which the linearized model was developed. It is clear fromthe figure that as the difference h-ho increases; the linearized approximation becomes

progressively less accurate.

A

B (linearized model)

ho

h

t

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Theoretical Models of Chemical Processes